A mass of 20 g stretches a spring 5 cm. Suppose that the mass is also attached to a viscous damper with a damping constant of 400 dyn·s/cm. If the mass is pulled down an additional 2 cm and then released, find its position u at any time t. Plot u versust. Determine the quasi frequency and the quasi period. Determine the ratio of the quasi period to the period of the corresponding undamped motion. Also find the time τ such that |u(t)| < 0.05 cm for all t > τ.

Quasi frequency = 4√6

Quasi period = π√6/12

t ≈ 0.4045

Explanation:

Given:

Mass, m = 20g

τ = 400 dyn.s/cm

k = 3920

u(0) = 2

u'(0) = 0

General differential equation:

mu" + τu' + ku = 0

Replacing the variables with the known value:

20u" + 400u' + 3920u = 0

Divide each side by 20

u" + 20u' + 196u = 0

Determining the characteristic equation by replacing y" with r², y' with r and y with 1 in the differential equation.

r² + 20r + 196 = 0

Determining the roots:

r = -10 ± 4√6i

The general solution for two complex roots are:

y = c₁ eᵃt cosbt + c₂ eᵃt sinbt

with a the real part of the roots and b be the imaginary part of the roots.

Since, a = -10 and b = 4√6

u(t) = c₁e⁻¹⁰^t cos 4√6t + c₂e⁻¹⁰^t sin 4√6t

u(0) = 2

u'(0) = 0

(b)

Quasi frequency:

μ =

(c)

Quasi period:

T = 2π / μ

(d)

|u(t)| < 0.05 cm

u(t) = |2e⁻¹⁰^t cos 4√6t + 5√6/6 e⁻¹⁰^t sin 4√6t < 0.05

solving for t:

τ = t ≈ 0.4045

report

explanation:

this could be newton#'s law of cooling. the cooling curve is exponential